Tubing Pressure drops with Compressible Fluids

[Dullard]

TL;DR

Is this a valid technique?

I am a poor, dumb EE often stuck with the odd plumbing calculation. I am often asked questions like: "what size tubing do I need to convey 10 SLPM of 20 PSIG oxygen 200' with no more than 2 PSI pressure drop?"

I generally treat the fluid as in-compressible and use Darcy-Weisbach (I like Bellos for friction factor). So long as the pressure drop is < 10% of the inlet absolute pressure, my calcs correspond pretty well with reality. I try to stay conservative (It's not usually obvious when a tube is a little bit too big).

I had a situation this week where my installers were certain that they knew what they needed (they didn't ask me anything). They screwed up - the tubing will have to be replaced. When I did my normal calculation on what they actually installed it was too small by my standards, but should have worked (just - if I ignored the 10% rule). This got me thinking that I'd like to be able to (semi-accurately) go a bit beyond my previous comfort zone. I created an excel workbook (with some automation) to break a tubing run into 'n' segments. The outlet conditions for one segment are the inlet conditions for the next. This (mathematically) appears to work pretty well: The calculated pressure drop increases with 'n' and converges (increases at a decreasing rate). I'd appreciate any comments on the accuracy/validity/limits of this approach. If I'm missing an alternative approach, I'd love to hear about that, too. Thanks.

 

[jrmichler]

That sounds exactly like the approach that I vaguely remember from undergrad fluids class. Typical design problems require only enough design calculations to make sure that a tube is big enough, but not grossly oversized, so high accuracy is rarely needed. Capillary tubes in air conditioning systems may have a tight flow resistance tolerance.

 

[Chestermiller]

I would approach this differently. From Darcy-Weisbach, we have $$-\frac{dp}{dx}=\frac{\rho v^2 }{2D}f(Re,\epsilon)$$where f is the friction factor. The velocity is related to the mass flow rate m by
$$v=\frac{4m}{\rho \pi D^2}$$So, combining these two equations, we have: $$-\frac{dp}{dx}=\frac{8m^2 }{\rho \pi^2 D^5}f(Re,\epsilon)$$and $$Re=\frac{4m}{\pi D\mu}$$where $\mu$ is the viscosity. Everything on the right hand side is constant, except the density. But, from the ideal gas law, we have: $$\rho=\frac{pM}{RT}$$where M is the molecular weight. Substituting this yields:$$-\frac{dp^2}{dx}=\frac{16RTm^2 }{M \pi^2 D^5}f(Re,\epsilon)$$Integrating this yields: $$p^2=p_0^2-\frac{16RTm^2L }{M \pi^2 D^5}f(Re,\epsilon)$$

 

[Dullard]

I was sure that an elegant solution was possible - I posted the question because I thought "Chestermiller will know how to do this." I follow what you did - but I never would have gotten there by myself. Thank you very much!

Jim